9d^2+18d+5=0

Solution for 9d^2+18d+5=0 equation:

9d^2+18d+5=0

a = 9; b = 18; c = +5;
Δ = b2-4ac
Δ = 182-4·9·5
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-12}{2*9}=\frac{-30}{18}
=-1+2/3
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+12}{2*9}=\frac{-6}{18}
=-1/3
$